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"""
Project Euler Problem 20: Factorial Digit Sum
=============================================
.. module:: solutions.problem20
:synopsis: My solution to problem #20.
The source code for this problem can be
`found here <https://bitbucket.org/nekedome/project-euler/src/master/solutions/problem20.py>`_.
Problem Statement
#################
:math:`n!` means :math:`n \\times (n - 1) \\times \\dots \\times 3 \\times 2 \\times 1`
| For example, :math:`10! = 10 \\times 9 \\times \\dots \\times 3 \\times 2 \\times 1 = 3628800`,
| and the sum of the digits in the number :math:`10!` is :math:`3 + 6 + 2 + 8 + 8 + 0 + 0 = 27`.
Find the sum of the digits in the number :math:`100!`
Solution Discussion
###################
Build the value of :math:`100!` using Python's arbitrary precision arithmetic and then perform a decimal digit sum on
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that result.
Solution Implementation
#######################
.. literalinclude:: ../../solutions/problem20.py
:language: python
:lines: 35-
from lib.digital import digit_sum
from lib.sequence import Factorials
def solve():
""" Compute the answer to Project Euler's problem #20 """
target = 100
factorials = Factorials()
x = factorials[target]
x
(([a-z][a-z0-9_]{2,30})|(_[a-z0-9_]*))$
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answer = digit_sum(x)
return answer
expected_answer = 648
expected_answer
(([A-Z_][A-Z0-9_]*)|(__.*__))$
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